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Why is the following property true for the Schouten tensor $P_{ij}$:

$$\nabla^k\nabla_iP_{kj}=\nabla^k\nabla_jP_{ki}$$

The Schouten tensor is defined as $$P_{ij}=\frac{1}{n-2}\left(\operatorname{Ric}_{ij}-\frac{1}{2(n-1)}Rg_{ij}\right)$$

I need this fact to prove that the Bach tensor is symmetric. But I don't know why the above fact should be true. I know that Bianchi identities imply that $\nabla^kP_{kj}=\nabla_jP^k_k$. But this fact is also not helping me prove the above identity.

Ryan Hendricks
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  • You need to use the formula for $\nabla_a\nabla_bP_{ij} - \nabla_b\nabla_aP_{ij}$ – Deane May 23 '22 at 00:55
  • @Deane- What is the formula for that? This is just $R_{ab}P_{ij}$ right? I think that the answer to this will be able to solve my problem. – Ryan Hendricks May 23 '22 at 01:04
  • @Deane I don't think this will work... the LHS is $\nabla^k\nabla_i$ whereas the right hand side is $\nabla^k\nabla_j$. – K.defaoite May 23 '22 at 01:06
  • @Deane Sorry, I see your idea now. My fault. – K.defaoite May 23 '22 at 01:09
  • @K.defaoite- What is the idea? – Ryan Hendricks May 23 '22 at 01:11
  • See https://math.stackexchange.com/a/3818375/10584 – Deane May 23 '22 at 01:18
  • @Deane- So we have $\nabla_a\nabla_b P_{ij}-\nabla_b \nabla_a P_{ij}=R_{abi}{}^cP_{cj}+R_{abj}{}^cP_{ic}$. I don't know how to proceed from here. If I take a trace over $a$ and $i$, I get $\nabla^k\nabla_b P_{kj}=Ric_b{}^cP_{cj}+R^k_{bj}{}^c P_{kc}+\nabla_b\nabla_j(tr_gP)$. I don't know how this would help. – Ryan Hendricks May 23 '22 at 01:46

1 Answers1

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Remember the Ricci Identity, $$[\nabla_a,\nabla_b]A_{c_1\dots c_n}=A_{d~c_2\dots c_n}R^d{}_{c_1ab}+A_{c_1d~c_3\dots c_n}R^d{}_{c_2ab}+\dots A_{c_1\dots c_{n-1}d}R^d{}_{c_nab}$$ And so,

$$[\nabla^k,\nabla_i]P_{kj}=P_{lj}R^l{}_{k}{}^k{}_{i}+P_{kl}R^l{}_{j}{}^k{}_{i}$$

Now, note that $R^l{}_k{}^k{}_{i}=R^{lk}{}_{ki}=-R^{kl}{}_{ki}=-R^l{}_{i}$

So,

$$[\nabla^k,\nabla_i]P_{kj}=-P_{lj}R^l{}_i+P_{kl}R^l{}_{j}{}^k{}_{i}$$

And so, expanding the commutator,

$$\nabla^k\nabla_iP_{kj}=\nabla_i\nabla^kP_{kj}-P_{lj}R^l{}_i+P_{kl}R^l{}_{j}{}^k{}_{i} \\ =\nabla_i\nabla_jP^k{}_{k}-P_{lj}R^l{}_i+P_{kl}R^l{}_{j}{}^k{}_{i}$$

And of course, swapping the roles of $i,j$ this gives

$$\nabla^k\nabla_jP_{ki} =\nabla_j\nabla_iP^k{}_{k}-P_{li}R^l{}_j+P_{kl}R^l{}_{i}{}^k{}_{j}$$

Combining, $$\nabla^k\nabla_iP_{kj}-\nabla^k\nabla_jP_{ki}=[\nabla_i,\nabla_j]P^k{}_k+(P_{li}R^l{}_j-P_{lj}R^l{}_i)+(P_{kl}R^l{}_{j}{}^k{}_{i}-P_{kl}R^l{}_{i}{}^k{}_{j}) \tag{*}$$

First term: Since $P^k{}_k$ is a scalar, $[\nabla_i,\nabla_j]P^k{}_k=0$.

Second term: Observe

$$P_{li}R^l{}_j=\frac{1}{n-2}R^l{}_j\left(R_{li}-\frac{1}{2(n-1)}g_{li}\right) \\ = \frac{1}{n-2}R^l{}_j R_{li}-\frac{1}{2(n-1)(n-2)}R_{ij}$$

Which is very obviously symmetric w.r.t $i,j$.

Last term: We can exploit the Riemann tensor identities here.

$$P_{kl}R^l{}_j{}^k{}_i=P^{kl}R_{lj~ki} \\ =P^{kl}R_{ki~lj}$$

Now using the symmetry of $P$,

$$P^{kl}R_{ki~lj}=P^{lk}R_{ki~lj}$$

$l,k$ are dummy indices so we can swap them:

$$P^{lk}R_{ki~lj}=P^{kl}R_{li~kj} \\ = P_{kl}R^l{}_i{}^k{}_j$$

So the third term of $(*)$ vanishes as well. Thus finally,

$$\boxed{\nabla^k\nabla_iP_{kj}=\nabla^k\nabla_jP_{ki}}$$

As desired.

K.defaoite
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